In an experiment on the specific heat of a metal, a $0.20\ \text{kg}$ block of the metal at $150\ ^\circ\text{C}$ is dropped in a copper calorimeter (of water equivalent $0.025\ \text{kg}$) containing $150\ \text{cm}^{3}$ of water at $27\ ^\circ\text{C}$. The final temperature is $40\ ^\circ\text{C}$. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

> [!TLDR] > Using the heat-balance equation, the specific heat of the metal comes out to about **$432.9\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$**. If heat is lost to the surroundings during the experiment, the water and calorimeter actually receive *less* heat than the metal truly gave up - so the computed specific heat is **smaller** than the true value. ## Setting up the heat balance Heat lost by the metal block equals heat gained by the water plus the calorimeter (its "water equivalent" means it behaves thermally like an extra mass of water): - Metal: $m_m = 0.20$ kg, cools from $150^\circ$C to $40^\circ$C, $\Delta T_m = 110^\circ$C. - Water: $150\ \text{cm}^3 \Rightarrow m_w = 0.150$ kg (density $1\ \text{g/cm}^3$), warms from $27^\circ$C to $40^\circ$C, $\Delta T_w = 13^\circ$C. - Calorimeter's water equivalent: $W = 0.025$ kg, also warms by $13^\circ$C. ## Solving for the specific heat $s$ $$m_m\,s\,\Delta T_m = (m_w+W)\,c_w\,\Delta T_w$$ $$0.20\times s\times110 = (0.150+0.025)\times4186\times13$$ $$22\,s = 0.175\times4186\times13 = 9523.15$$ $$s = \frac{9523.15}{22} \approx432.9\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}.$$ ## Effect of heat loss to the surroundings This calculation assumes *all* the heat given up by the metal ends up in the water and calorimeter. If, in reality, some heat escapes to the surroundings during the experiment, then the water and calorimeter receive **less** heat than the metal actually released. But we only measure the water's temperature rise - from that (smaller, real) rise we compute $s$ as if it represented the *entire* heat the metal lost. …

In an experiment on the specific heat of a metal, a 0.20kg block of the metal at 150circC is dropped in a copp

Exercises · 10.14

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Concept understanding — Specific Heat Capacity

What is Specific Heat Capacity?

Imagine you have two identical stoves, two identical pots, and you put 1 kg of water in one pot and 1 kg of iron in the other. You turn both stoves to the same flame. After 2 minutes, the iron is scorching hot — you can't touch it. The water is still lukewarm.

Why? Because different substances need different amounts of heat to raise their temperature by the same amount. That's the core idea behind specific heat capacity.

The Intuition

Think of heat as "energy currency" and temperature rise as "buying a degree." Some materials are "cheap" — a little heat buys a big temperature rise. Others are "expensive" — you need to spend a lot of heat to get even a small rise.

Iron is cheap: a small heat input → large temperature jump.
Water is expensive: a large heat input → small temperature jump.

This "expensiveness" is what we call specific heat capacity. It tells you how much heat energy is needed to raise the temperature of 1 kg of a substance by 1 °C (or 1 K).

The Precise Definition

c = \frac{Q}{m \Delta T}

Where:

$c$ = specific heat capacity (J/kg·°C or J/kg·K)
$Q$ = heat energy supplied (J)
$m$ = mass of the substance (kg)
$\Delta T$ = change in temperature (°C or K)

In words: Specific heat capacity is the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin).

Key Points to Remember

It's a property of the material, not the object. A small iron nail and a giant iron beam have the same $c$ value — but the beam needs more total heat because it has more mass.
Units matter. Common values:
- Water: $c = 4186 \text{ J/kg·°C}$ (or ≈ 4200 J/kg·°C in many problems)
- Iron: $c \approx 450 \text{ J/kg·°C}$
- Copper: $c \approx 390 \text{ J/kg·°C}$
Notice water's value is about 10 times that of iron — that's why water heats up so slowly compared to metals.
The formula works both ways. If a substance cools down, it releases the same amount of heat it would absorb to warm up by the same $\Delta T$ .

A Common Mistake to Avoid

Watch out

Don't confuse specific heat capacity ( $c$ ) with heat capacity ( $C$ ). Heat capacity is for an entire object: $C = mc$ . A large iron block can have a higher heat capacity than a small cup of water, even though iron's $c$ is much smaller. Always check: are we talking about per kg or for the whole thing?

Worked Example (Exam-Style)

Problem: How much heat is needed to raise the temperature of 2 kg of water from 20 °C to 50 °C? (Take $c_{\text{water}} = 4200 \text{ J/kg·°C}$ )

Solution:

$m = 2 \text{ kg}$
$\Delta T = 50 - 20 = 30 \text{ °C}$
$c = 4200 \text{ J/kg·°C}$

Q = mc\Delta T = 2 \times 4200 \times 30 = 252\,000 \text{ J} = 252 \text{ kJ}

Answer: 252 kJ of heat is required.

Why This Matters …

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Using the heat-balance equation, the specific heat of the metal comes out to about $432.9\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$ . If heat is lost to the surroundings during the experiment, the water and calorimeter actually receive less heat than the metal truly gave up - so the computed specific heat is smaller than the true value.

Setting up the heat balance

Heat lost by the metal block equals heat gained by the water plus the calorimeter (its "water equivalent" means it behaves thermally like an extra mass of water):

Metal: $m_m = 0.20$ kg, cools from $150^\circ$ C to $40^\circ$ C, $\Delta T_m = 110^\circ$ C.
Water: $150\ \text{cm}^3 \Rightarrow m_w = 0.150$ kg (density $1\ \text{g/cm}^3$ ), warms from $27^\circ$ C to $40^\circ$ C, $\Delta T_w = 13^\circ$ C.
Calorimeter's water equivalent: $W = 0.025$ kg, also warms by $13^\circ$ C.

Solving for the specific heat $s$

m_m\,s\,\Delta T_m = (m_w+W)\,c_w\,\Delta T_w

0.20\times s\times110 = (0.150+0.025)\times4186\times13

22\,s = 0.175\times4186\times13 = 9523.15

s = \frac{9523.15}{22} \approx432.9\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}.

Effect of heat loss to the surroundings

This calculation assumes all the heat given up by the metal ends up in the water and calorimeter. If, in reality, some heat escapes to the surroundings during the experiment, then the water and calorimeter receive less heat than the metal actually released. But we only measure the water's temperature rise - from that (smaller, real) rise we compute $s$ as if it represented the entire heat the metal lost. …

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Setting up the heat balance

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Effect of heat loss to the surroundings

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Solving for the specific heat $s$

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